1. Oblicz ile moli stanowi 216g wody
2. Oblicz objętość w dm3 jaką będzie miało 88g tlenku węgla(IV)
masa molowa H2O = 18g/mol
216/18 = 12 moli
44g CO₂ ---- 22,4 dm³88g CO₂ ---- xdm³
x = 44,8dm³ CO₂
Zad1
ms=216g
M(H₂O)=18g/mol
n=ms/M
n=216g / 18g/mol = 12moli
Zad2
M(CO₂)=44g/mol
44g --- 22,4dm³
88g --- x
x=88g·22,4dm³/44g=44,8dm³
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masa molowa H2O = 18g/mol
216/18 = 12 moli
44g CO₂ ---- 22,4 dm³
88g CO₂ ---- xdm³
x = 44,8dm³ CO₂
Zad1
ms=216g
M(H₂O)=18g/mol
n=ms/M
n=216g / 18g/mol = 12moli
Zad2
M(CO₂)=44g/mol
44g --- 22,4dm³
88g --- x
x=88g·22,4dm³/44g=44,8dm³